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c++ - 如何将 time-uuid(存储在 boost uuid 中)转换为自纪元以来的时间戳/时间?

转载 作者:行者123 更新时间:2023-11-30 04:42:58 25 4
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从 EPOCH 开始从 UUID 时间戳转换为秒数似乎很容易基于 on the specs , 也基于 Cassandra's C++ driver source code基于 its struct definition .

但是,当我尝试这样做时,我总是得到错误的值。我做错了什么,我无法弄清楚它是什么。

为此,我使用了 here 提供的示例 UUID 值和 here .

所有要做的就是从 UUID 原始数据中取出第一个 uint64_t,屏蔽它的前四个 MSb,减去一个差值并除以一个数。

这是我的最小完整示例:

#include <boost/date_time.hpp>
#include <boost/uuid/uuid.hpp>
#include <boost/uuid/uuid_generators.hpp>
#include <boost/uuid/uuid_io.hpp>
#include <cstdint>
#include <iostream>

uint64_t TimestampFromUUID(const boost::uuids::uuid& uuid) {
static constexpr const int UUID_SIZE = 16;
static_assert(sizeof(uuid) == UUID_SIZE, "Invalid size of uuid");

static constexpr const int MS_FROM_100NS_FACTOR = 10000;
static constexpr const uint64_t OFFSET_FROM_15_10_1582_TO_EPOCH = 122192928000000000;

struct two64s {
uint64_t n1;
uint64_t n2;
} contents;
std::memcpy(&contents, uuid.data, UUID_SIZE);
// contents.n1 = __builtin_bswap64(contents.n1);
uint64_t timestamp = contents.n1 & UINT64_C(0x0FFFFFFFFFFFFFFF);
return (timestamp - OFFSET_FROM_15_10_1582_TO_EPOCH) / MS_FROM_100NS_FACTOR;
}

int main() {
std::cout << "Time now: " << (boost::posix_time::second_clock::universal_time() - boost::posix_time::ptime(boost::gregorian::date(1970, 1, 1))).total_milliseconds() << std::endl;
auto gen = boost::uuids::string_generator();
std::cout << "UUID: " << gen("49cbda60-961b-11e8-9854-134d5b3f9cf8") << std::endl;
std::cout << "Time from UUID: " << TimestampFromUUID(gen("49cbda60-961b-11e8-9854-134d5b3f9cf8")) << std::endl;
std::cout << "UUID: " << gen("58e0a7d7-eebc-11d8-9669-0800200c9a66") << std::endl;
std::cout << "Time from UUID: " << TimestampFromUUID(gen("58e0a7d7-eebc-11d8-9669-0800200c9a66")) << std::endl;

return 0;
}

这个程序的输出是:

Time now: 1571735685000
UUID: 49cbda60-961b-11e8-9854-134d5b3f9cf8
Time from UUID: 45908323159150
UUID: 58e0a7d7-eebc-11d8-9669-0800200c9a66
Time from UUID: 45926063291384

你可以玩这个源代码here .

为什么我的结果甚至不接近当前时间戳?我做错了什么?

最佳答案

我觉得把UUID处理成字符串,用字符串操作提取时间戳信息,再转换成数值会更容易理解。诀窍在于时间戳信息存储在 UUID 中的方式。来自规范:

The formal definition of the UUID string representation is provided by the following ABNF [7]:

  UUID                   = time-low "-" time-mid "-"
time-high-and-version "-"
clock-seq-and-reserved
clock-seq-low "-" node
time-low = 4hexOctet
time-mid = 2hexOctet
time-high-and-version = 2hexOctet
clock-seq-and-reserved = hexOctet
clock-seq-low = hexOctet
node = 6hexOctet
hexOctet = hexDigit hexDigit
hexDigit =
"0" / "1" / "2" / "3" / "4" / "5" / "6" / "7" / "8" / "9" /
"a" / "b" / "c" / "d" / "e" / "f" /
"A" / "B" / "C" / "D" / "E" / "F"

The following is an example of the string representation of a UUID as a URN:

urn:uuid:f81d4fae-7dec-11d0-a765-00a0c91e6bf6

即UUID 的第一部分(在“-”之前)是低版本,第二部分是中版本,第三部分是高版本,第一个字符是 UUID 版本。所以我们需要拆分 UUID 并重新组合这些时间戳部分以创建完整的时间戳字符串,如下所示:{time-high minus-version}{time-mid}{time-low}

这是修改后的代码。我将这个不错的 javascript 示例作为引用:https://stackoverflow.com/a/26915856/3694234

#include <boost/date_time.hpp>
#include <boost/uuid/uuid.hpp>
#include <boost/uuid/uuid_generators.hpp>
#include <boost/uuid/uuid_io.hpp>
#include <boost/algorithm/string.hpp>
#include <string>
#include <vector>
#include <cstdint>
#include <iostream>

uint64_t TimestampFromUUID(const boost::uuids::uuid& uuid) {
static constexpr const int UUID_SIZE = 16;
static_assert(sizeof(uuid) == UUID_SIZE, "Invalid size of uuid");

static constexpr const int MS_FROM_100NS_FACTOR = 10000;
static constexpr const uint64_t OFFSET_FROM_15_10_1582_TO_EPOCH = 122192928000000000;

/* convert uuid to string for manipulation */
std::string uuid_str = boost::uuids::to_string(uuid);
/* store uuid parts in a vector */
std::vector<std::string> uuid_parts;

/* split uuid with '-' as delimiter */
boost::split(uuid_parts, uuid_str, [](char c){return c == '-';});

/* first part of uuid is time-low
second part is time-mid
third part is time high with most significant 4 bits as uuid version
*/
std::string uuid_timestamp = uuid_parts[2].substr(1) + uuid_parts[1] + uuid_parts[0];
std::cout << std::endl << "UUID Timestamp : " << uuid_timestamp << std::endl;

uint64_t timestamp = std::stoul(uuid_timestamp, nullptr, 16);

return (timestamp - OFFSET_FROM_15_10_1582_TO_EPOCH) / MS_FROM_100NS_FACTOR;
}

int main() {
std::cout << "Time now: " << (boost::posix_time::second_clock::universal_time() - boost::posix_time::ptime(boost::gregorian::date(1970, 1, 1))).total_milliseconds() << std::endl;
auto gen = boost::uuids::string_generator();
std::cout << "UUID: " << gen("49cbda60-961b-11e8-9854-134d5b3f9cf8") << std::endl;
std::cout << "Time from UUID: " << TimestampFromUUID(gen("49cbda60-961b-11e8-9854-134d5b3f9cf8")) << std::endl;
std::cout << "UUID: " << gen("58e0a7d7-eebc-11d8-9669-0800200c9a66") << std::endl;
std::cout << "Time from UUID: " << TimestampFromUUID(gen("58e0a7d7-eebc-11d8-9669-0800200c9a66")) << std::endl;

return 0;
}

输出

Time now: 1571838175000
UUID: 49cbda60-961b-11e8-9854-134d5b3f9cf8
Time from UUID:
UUID Timestamp : 1e8961b49cbda60
1533190458118
UUID: 58e0a7d7-eebc-11d8-9669-0800200c9a66
Time from UUID:
UUID Timestamp : 1d8eebc58e0a7d7
1092575371981

关于c++ - 如何将 time-uuid(存储在 boost uuid 中)转换为自纪元以来的时间戳/时间?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/58500967/

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